TS EAMCET · Maths · Application of Derivatives
Let \(P(x)\) be a polynomial of degree 3 having extreme value at \(x=1\). If \(\lim _{x \rightarrow 0}\left(\frac{P(x)+4}{x^2}+2\right)=6\), then \(\left(\frac{d P}{d x}\right)_{x=\frac{1}{2}}=\)
- A 2
- B 0
- C -2
- D 4
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { (a) Let } P(x)=a x^3+b x^2+c x+d \\ & \therefore \quad P^{\prime}(x)=3 a x^2+2 b x+c \end{aligned}\) Since, \(P(x)\) has extreme value at \(x=1\) \(\therefore \quad P^{\prime}(1)=0 \Rightarrow 3 a+2 b+c=0\) ...(i) Now, we have…
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